Is turning off MOSFET too fast, bad?
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked 4 hours ago
Sudoer
1278
add a comment |
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked 4 hours ago
Sudoer
1278
add a comment |
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked 4 hours ago
Sudoer
1278
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
mosfet driver
asked 4 hours ago
Sudoer
1278
asked 4 hours ago
Sudoer
1278
asked 4 hours ago
Sudoer
1278
asked 4 hours ago
Sudoer
1278
asked 4 hours ago
Sudoer
1278
1278
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
answered 3 hours ago
Sparky256
11k21534
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414759%2fis-turning-off-mosfet-too-fast-bad%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
answered 3 hours ago
Sparky256
11k21534
add a comment |
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
answered 3 hours ago
Sparky256
11k21534
add a comment |
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
answered 3 hours ago
Sparky256
11k21534
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
answered 3 hours ago
Sparky256
11k21534
edited 23 mins ago
answered 3 hours ago
Sparky256
11k21534
answered 3 hours ago
Sparky256
11k21534
answered 3 hours ago
Sparky256
11k21534
11k21534
add a comment |
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414759%2fis-turning-off-mosfet-too-fast-bad%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
